Question

What is the digit in the units place of the sum of 1^1+ 2^2+ 3^3+ 4^4 +.....+ 99^99 + 100^100?

Answer 1

Let us write down first few factors

1^1 = 1

2^2 = 4

3^3 = 27

4^4 = 256

5^5 = 3125

6^6 = 46656

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100^100 = ... finish in zero

The last two digits in the sum would be 20

The digit in the unit would be 0